4 4 dt et 2 sin2 4 integratingdt factor e t 21 sin2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sin(2 t) . ( (a) The a commodity eD 1+cos(2 t = e demand t)20+4p)) by 5.6.(Consider DE is Factor: (withS ) = dt)(20+8p+ 1p3/2 curves given= (4p p3/2 ), with IC p(0) = p0 . 4 4 dt = et 2 sin(2 . 4 Integratingdt Factor: e t+ 21 sin(2 t) dp t 2p Equilibrium solutions:(1 p) cos(2⇡20 ++41 sp(4 t) p p1/2 ) 5(1 + = 0, 16. et+ 21 sin(2 t) 4+ p ) p cos(2⇡ t)) S ( p =3/2 t= e0 ) in(2 e + )) = 1 1 0 3 0 dt Bernoulli+equationdp D(pn +=3/20⇡Let tv 2=3in(2 3t/2 == 1/2 ) cos(2⇡ t))pt+ /12 pin(2Using the 2. t 8e 1 p1 et 2 sin(2 t) with(1 =cos(2 +))p + p s/2 ) p p 5(1 + v = 1 2 e 2 s . t) +) d t+ 1 sin(2 t) dt DE: e2 p = 5(1 + cos(2⇡ t))et+ 2 sin(2 t) dt d rate of sin(2 t) 1 The rate of increase of price is 1/4 the t+e1t1sin(2 decrease = 1quantity and t))et+ 21 sin(2 t) +2 1 of 5(1 +) cos(2⇡ the t) p t+ sin(2 t + C v 0 = e 2 p 1/2p+ = 5e 2 dt initial price is p0 . ) 1 2 in(2 General Solution:21psin(2 = p 5 = C 5et +221 ssin(2 t)t) + C t8 et et+ (1) t) 1 a) (Set up an initial value problem = that models thisLinear DE and find all situation 1 v+ IC: p(0) = p ) p0 = 5 C )General Solution: p(t) = 5 C e t 2 sin(2 t) C = 5 p0 . equilibrium 0 solutions of the differential2in(2 t) 8 equation. Then find the solution 1 Solution of IVP: p(t) = 5 (5 p )e t 2 s . 1 of the(0) =form: v 00+ lim = t1 .= 50C = 5 p0 . solutions:2 dt(t) =t/2 . IC: p initial 0value problem. ) Integrating factor: e 1/ p = e 5. Standard behaviour: =25 p(C ) . Equilibrium v8 Longterm p ) p ) t Solution: IVP: p(t) = 5 (5 p0 )e t 21 sin(2 t) . Solution of 1 t/2 /2 1 t/2 dp 1 t/2 1 0 = (20+4p)) = 1 (4p p / t = 5. e 6. (a) Longterm behaviour: Slimvp(t) = 5.v p3Equilibrium solutions: 3p2(),)with IC p(0) = p0 . The DE is = (D e) = + e p (20+8 t 8 dt 4 42 4 Equilibriumdp solutions: 4p p3/21= t/2 p(4 p1/2 )2) p = 0, 16. 1 d 0) 1 t/ 1 Bernoulli equation with n ) =dt [e Let ]vp= p1 82e 2 (20+42 )) =0 = p1 p p3/22p0 . with IC p(0) = p0 . 6. (a) The DE is = (D S = 3/2. v = 3/ 3/ = p 1/ p ) v (4 2 3/ ), Using the (20+8 dt 4 4 4 DE: Equilibrium solutions: 4p p3/2et/2 v ) p(4 et/2 1/2 C) p = 0, 16. = 0 = 1 p+ ) 1 0 4 Bernoulli equation with n v= 3/2. 1 p 1/2 + 1p1 3/2 = p...
View Full Document

Ask a homework question - tutors are online